Ta có: \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow\left(x-y\right)^2\ge0\) (đúng)
Áp dụng ta có:\(x^4+y^4\ge\frac{\left(x^2+y^2\right)^2}{2}\ge\frac{\left(\frac{\left(x+y\right)^2}{2}\right)^2}{2}=\frac{\left(\frac{1}{2}\right)^2}{2}=\frac{\frac{1}{4}}{2}=\frac{1}{8}\)
Vậy \(Min_M=\frac{1}{8}\) khi \(x=y=\frac{1}{2}\)
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